Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
/*
 * 2608.cc: Minus One
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 100000;
const int INF = 1 << 30;

/* typedef */

typedef long long ll;
typedef vector<int> vi;
typedef queue<int> qi;

/* global variables */

vi nbrs[MAX_N];
int dists0[MAX_N], dists1[MAX_N], cnts0[MAX_N], cnts1[MAX_N];

/* subroutines */

void dijkstra(int n, int dists[], int cnts[], int st, int gl) {
  for (int i = 0; i < n; i++) dists[i] = INF;
  dists[st] = 0;

  qi q;
  q.push(st);

  while (! q.empty()) {
    int u = q.front(); q.pop();
    int ud = dists[u];
    cnts[ud]++;
    if (u == gl) break;

    int vd = ud + 1;
    vi &nbru = nbrs[u];
    for (vi::iterator vit = nbru.begin(); vit != nbru.end(); vit++) {
      int &v = *vit;
      if (dists[v] > vd) {
	dists[v] = vd;
	q.push(v);
      }
    }
  }
  //for (int i = 0; i <= dists[gl]; i++) printf("%d: %d\n", i, cnts[i]);
}

/* main */

int main() {
  int n, m, st, gl;
  cin >> n >> m >> st >> gl;
  st--, gl--;

  for (int i = 0; i < m; i++) {
    int xi, yi;
    cin >> xi >> yi;
    xi--, yi--;
    nbrs[xi].push_back(yi);
    nbrs[yi].push_back(xi);
  }

  dijkstra(n, dists0, cnts0, st, gl);
  dijkstra(n, dists1, cnts1, gl, st);
  
  ll sum = 0;
  int dg = dists0[gl];
  for (int i = 0; i <= dg - 2; i++)
    sum += (ll)cnts0[i] * cnts1[dg - i - 2];
  printf("%lld\n", sum);
  return 0;
}